We know that $0<\dfrac{n^2-2}{n^6+1}<\dfrac{n^2-2}{n^6}<\dfrac{n^2}{n^6}=\dfrac{1}{n^4}$ for any $n\ge 2$. Considering this fact, what does the direct comparison test say about $\sum_{n=2}^{\infty }{\frac{n^2-2}{{{n}^{6}}+1}}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A The series converges. (Choice B) B The series diverges. (Choice C) C The test is inconclusive.
Answer: $\sum_{n=2}^{\infty }~{\frac{1}{{{n}^{4}}}}~~$ is a $~p$ -series with $~p=4~$, so it converges. Our given series is term-by-term less than a convergent series, so it converges as well.